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<record version="4" id="5567">
 <title>group scheme of multiplicative units</title>
 <name>GroupSchemeOfMultiplicativeUnits</name>
 <created>2004-02-11 04:19:26</created>
 <modified>2005-03-25 20:39:15</modified>
 <type>Example</type>
<parent id="3122">scheme</parent>
 <creator id="2727" name="mathcam"/>
 <author id="2727" name="mathcam"/>
 <author id="4430" name="archibal"/>
 <classification>
	<category scheme="msc" code="14A15"/>
 </classification>
 <defines>
	<concept>group scheme of multiplicative units</concept>
 </defines>
 <synonyms>
	<synonym concept="group scheme of multiplicative units" alias="$\mathbb{G}_m$"/>
 </synonyms>
 <related>
	<object name="GroupScheme"/>
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 <content>\PMlinkescapeword{sort}
Let $R=\mathbb{Z}[X,Y]/\left&lt;XY-1\right&gt;$.  Then $\Spec R$ is an affine scheme.  The natural homomorphism $\mathbb{Z}\to R$ makes $R$ into a scheme over $\Spec \mathbb{Z}$, i.e. a $\mathbb{Z}$-scheme.  

What are the $\mathbb{Z}$-points of $\Spec R$? Recall that an $S$-point of a scheme  $X$ is a morphism $S\to X$; if we are working in the category of schemes over $Y$, then the morphism is expected to commute with the structure morphisms.  So, here, we seek homomorphisms $\mathbb{Z}[X,Y]/\left&lt;XY-1\right&gt; \to \mathbb{Z}$. Such a homomorphism must take $X$ to an invertible element, and it must take $Y$ to its inverse.  Therefore there are two, one taking $X$ to $1$ and one taking $X$ to $-1$. One recognizes these as the multiplicative units of $\mathbb{Z}$, and indeed if $S$ is any ring, then the $S$-points of $\Spec R$ are exactly the multiplicative units of $S$.  For this reason, this scheme is often denoted $\mathbb{G}_m$. It is an example of a group scheme. 

We can regard any morphism as a family of schemes, one for each fibre.
Since we have a morphism $\mathbb{G}_m \to \mathbb{Z}$, we can ask about the fibres of this morphism.  If we select a point $x$ of $\Spec \mathbb{Z}$, we have two choices. Such a point must be a prime ideal of $\mathbb{Z}$, and there are two kinds: ideals generated by a prime number, and the zero ideal. If we select a point $x$ with residue field $k(x)$, then the fiber of this morphism will be $\Spec R \times \Spec k(x)$, which is the same as $\Spec R\otimes k(x)$. But looking at the definition of $R$, we see that this is $\Spec k(x)[X,Y]/\left&lt;XY-1\right&gt;$, which is just the scheme whose points are the nonzero elements of $k(x)$.  

In other words, we have a family of schemes, one in each characteristic.  Of course, normally one wants a family to have some additional sort of smoothness condition, but this demonstrates that it is quite possible to have a family of schemes in different characteristics; sometimes one can deduce the behaviour in one characteristic from the behaviour in another.  This approach can be useful, for example, when dealing with Hilbert modular varieties.</content>
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